Q 12-01-021NEETJEE MainMedium
A charge of $+1\ \mu$C is at the origin and $-1\ \mu$C is at $(0, 2\ \text{mm})$. The electric field at $(10\ \text{cm}, 0)$ is approximately \left(\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9\ \text{N m}^2\text{C}^{-2}\right)
Answer: (A) $1.8 \times 10^4$ N/C
Concept. The pair is a short dipole and the point is on its equatorial line: $E = \dfrac{kp}{r^3}$.
Step 1. Dipole moment. $p = 10^{-6} \times 2 \times 10^{-3} = 2 \times 10^{-9}$ C m.
Step 2. Field. $r = 0.1$ m:
$$E = \frac{9 \times 10^9 \times 2 \times 10^{-9}}{10^{-3}} = 1.8 \times 10^4\ \text{N/C}$$
Answer. $1.8 \times 10^4$ N/C (Option A).
Solution by Sreeraj P, M.Sc Physics